A first order reaction has a rate constant `1.15xx10^(-3)s^(-1)`. How long will `5g` of this reactant take to reduce to `3g`` ?

Explanation

To solve the problem, we will use the formula for the time taken in a first-order reaction:

\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]

Where:
- \( t \) is the time taken,
- \( k \) is the rate constant,
- \( A \) is the initial amount of the reactant,
- \( x \) is the amount of reactant that has reacted.

### Step-by-step Solution:

1. **Identify the given values**:
- Rate constant \( k = 1.15 \times 10^{-3} \, s^{-1} \)
- Initial amount \( A = 5 \, g \)
- Final amount \( A - x = 3 \, g \)

2. **Calculate the amount reacted**:
- The amount reacted \( x = A - (A - x) = 5 \, g - 3 \, g = 2 \, g \)

3. **Substitute the values into the formula**:
\[ t = \frac{2.303}{1.15 \times 10^{-3}} \log \left( \frac{5}{3} \right) \]

4. **Calculate the logarithm**:
- Calculate \( \frac{5}{3} = 1.6667 \)
- Find \( \log(1.6667) \approx 0.2218 \)

5. **Calculate the time**:
\[ t = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2218 \]
\[ t = \frac{2.303 \times 0.2218}{1.15 \times 10^{-3}} \]
\[ t \approx \frac{0.5113}{1.15 \times 10^{-3}} \approx 444.0 \, s \]

6. **Final answer**:
- The time taken for the reactant to reduce from 5g to 3g is approximately **444 seconds**.